Question

1. Let A1, A2,..., An be mutually disjoint events. Show that

a) IP(A1UA2U...UAn) = IP(A1) + IP(A2) + ... + IP(An)

b) There exists at least one i with IP(Ai) less than equals to 1/n

2. Define conditional probability IP(E|F). Derive the Law of total probability and use it to derive Bayes's Formula

Answer 1

We would be looking at each part of the first question here.

Q1) a) The union probability here is computed as:
P(A₁ A₂ A₃ ...... A_n )  

= Probability that any of the events out of A₁, A₂, A₃, ..... A_n can happen

As all of these are mutually disjoint events, which is that they cannot happen together. Therefore the probability of having any of the event out of the n events is equal to the sum of probabilities of each of the n events here.

Using additivity law of probability, the intersection probability terms would cancel out as they all are mutually disjoint.

Therefore P(A₁ A₂ A₃ ...... A_n ) = P(A₁) + P(A₂) + .... + P(A_n) is the required expression here.

b) As there are n events, and all of them are mutually disjoint

Therefore none of them can happen together which means that the sum of all those probabilities of events happening individually is equal to the probability of any of the event happening ( as shown in the previous part )

Now as we know that each probability can only have a value less than or equal to 1. Therefore,

P(A₁ A₂ A₃ ...... A_n ) <= 1

P(A₁) + P(A₂) + .... + P(A_n) <= 1

Let us assume that each of the above probability is greater than 1/n

That would mean:

P(A₁) + P(A₂) + .... + P(A_n) > 1 because n*(1/n) = 1

Therefore the our assumption was incorrect, and we have to conclude here that there exist some P(A_i ) which is less than or equal to 1/n